Solve x^2 + 6x + 4=0?
1 answer:
You need to use the quadratic formula for this question.
In your equation, a is 1, b is 6, and c is 4. (I couldn't find a plus minus sign, so that's what the +- means, sorry if there's confusion.)
Sub those into the quadratic formula, and
For the first answer, we should get
x1=-0.7639, when you add the √6^2-4(1)(4)
For the second answer, we get
x2=-5.2361 when we subtract the √6^2-4(1)(4)
Therefore, the x values that make this equation equal to 0 should be approximately -0.7639 and -5.2361.
The more specific answers are
x1= -3+√5 and x2=-3-√5.
In your equation, a is 1, b is 6, and c is 4. (I couldn't find a plus minus sign, so that's what the +- means, sorry if there's confusion.)
Sub those into the quadratic formula, and
For the first answer, we should get
x1=-0.7639, when you add the √6^2-4(1)(4)
For the second answer, we get
x2=-5.2361 when we subtract the √6^2-4(1)(4)
Therefore, the x values that make this equation equal to 0 should be approximately -0.7639 and -5.2361.
The more specific answers are
x1= -3+√5 and x2=-3-√5.
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Answer:
Step-by-step explanation:
It is given that the point P(x, y) lies on the unit circle and is defined by real number e. We have to find csc(e) i.e cosecant of (e)
This scenario is represented in the figure attached below.
The formula of csc is:
A right angled triangle is constructed, as shown in figure. The hypotenuse of the triangle forms the radius of the circle. Since, the circle is a unit circle the length of hypotenuse would be equal to 1.
Hypotenuse = 1
So, the formula of csc would be reduced to:
The side opposite to the angle made at origin is y. So,
